from the given problem we can see that when the switch is open we have a series LRC circuit involving just the one
   caopacitor near the upper right corner so that we get
   [ω_d L - (1 / ω_d C)] / R = tanφ_o
                                       = tan (- 20^o)
                                       = - tan20^o

   now when the switch is in position 1 the equivalent capacitance in the circuit is 2C in this case we get that
   [ω_d L - (1 / 2 ω_d C)] / R = tanφ₁
                                       = tan (10.0^o)
   finally when the switch is in position 2 the circuit is an LC circuit with current amplitude
   I₂ = ε_m / Z_LC
       = ε_m / √[ω_d L - (1 / ω_d C)]² 
       = ε_m / [(1 / ω_d C) - ω_d L]
   where we use the fact that
   (ω_d C)^-1 > ω_d L while simplifying the square root
(a)
   the resistance can be found from
   R = - ε_m / I₂ tanφ_o
      = ....... Ω
(b)
   the inductance can be found from
   R = (ε_m / ω_d I₂) [1 - 2 (tanφ₁ / tanφ_o)]
      = ....... H
(c)
   C = I₂ / 2 ω_d ε_m (1 - (tanφ₁ / tanφ_o))
       = ............. F